Problem: Let $g(x)=\dfrac{2}{x^3}-\dfrac{1}{x^2}$. $g'(2)=$
The strategy We can first rewrite each rational term of $g$ as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Once we have $g'(x)$, we can plug $x=2$ into it to find $g'(2)$. Rewriting rational terms as negative powers $\begin{aligned} g(x)&=\dfrac{2}{x^3}-\dfrac{1}{x^2} \\\\ &=2x^{-3}-x^{-2} \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}(2x^{-3}-x^{-2}) \\\\ &=2\dfrac{d}{dx}(x^{-3})-\dfrac{d}{dx}(x^{-2}) \\\\ &=2(-3x^{-4})-(-2)x^{-3} \\\\ &=-6x^{-4}+2x^{-3} \\\\ &=-\dfrac{6}{x^4}+\dfrac{2}{x^3} \end{aligned}$ Evaluating $g'(x)$ So we found that $g'(x)=-\dfrac{6}{x^4}+\dfrac{2}{x^3}$. Let's plug $x=2$ into $g'$ : $\begin{aligned} &\phantom{=}g'(2) \\\\ &=-\dfrac{6}{(2)^4}+\dfrac{2}{(2)^3} \\\\ &=-\dfrac{6}{16}+\dfrac{2}{8} \\\\ &=-\dfrac{3}{8}+\dfrac{2}{8} \\\\ &=-\dfrac18 \end{aligned}$ In conclusion, $g'(2)=-\dfrac18$.